17.6 Substitution in Integrals

Let $f$ be a nice function on an interval $[a,b]$. Then if $F$ is any antiderivative for $f$, we have

$$\int_a^b f=F\mid_a^b =F(b)-F(a),$$

by the fundamental theorem of calculus. We saw in (17.32) that under suitable hypotheses on $g$, $F\circ g$ is an antiderivative for $(f\circ g)g'$. Hence

$$\int_a^b f\Big(g(t)\Big)g'(t)dt=F\circ g\mid_a^b=F\Big\vert _{g(a)}^{g(b)}=\int_{g(a)}^{g(b)}f(u)du.$$

Hence we can find $${\int_a^b f\Big( g(t)\Big)g'(t)dt}$$ by the following ritual:

Let $u=g(t)$. When $t=a$ then $u=g(a)$ and when $t=b$ then $u=g(b)$. Also $du=g'(t)dt$. Hence

$$\int_a^b f\Big( g(t)\Big) g'(t)dt=\int_{g(a)}^{g(b)} f(u)du=F(u)\Big\vert _{g(a)}^{g(b)}.$$

17.51   Example. To find $${\int_{\pi^2}^{4\pi^2}{{\sin(\sqrt x)}\over {\sqrt x}}dx}$$.

Let $u=\sqrt x$. When $x=\pi^2$, then $u=\pi$, and when $x=4\pi^2$, then $u=2\pi$. Also $${du={1\over {2\sqrt x}}dx}$$, so

$$\begin{eqnarray*} \int_{\pi^2}^{4\pi^2}{{\sin (\sqrt x)}\over {\sqrt x}}dx &=& 2... ...pi^{2\pi} \\ &=& -2\Big( \cos(2\pi)-\cos (\pi)\Big)=-2(1+1)=-4. \end{eqnarray*}$$

We saw in (17.36) that if $h$ is an inverse function for $g$, then an antiderivative for $f\circ g$ is $H\circ g$, where $H$ is an antiderivative for $f\cdot h'$. Thus

$$\int_a^b f\Big( g(t)\Big)dt =H\circ g\mid_a^b=H\Big\vert _{g(a)}^{g(b)}.$$

The ritual for finding $${ \int_a^b f\Big(g(t)\Big)dt}$$ in this case is:

Let $u=g(t)$. Then $t=h(u)$ and $dt=h'(u)du$. When $t=a$ then $u=g(a)$, and when $t=b$ then $u=g(b)$. Thus

$$\int_a^b f\Big( g(t)\Big)dt =\int_{g(a)}^{g(b)}f(u)h'(u)du=H(u)\Big\vert _{g(a)}^{g(b)}$$

where $H$ is an antiderivative for $fh'$.

17.52   Example. To find $${\int_0^{\ln(\sqrt 3)}{1\over {e^x+e^{-x}}}dx}$$.

Let $u=e^x$. When $x=0$ then $u=1$, and when $x=\ln(\sqrt 3)$ then $u=\sqrt 3$. Also $x=\ln (u)$, so $${ dx={1\over u}du}$$.

$$\begin{eqnarray*} \int_0^{\ln(\sqrt 3)}{1\over {e^x+e^{-x}}}dx &=& \int_1^{\sqrt... ...)-\arctan (1) \\ &=& {\pi\over 3}-{\pi\over 4}={\pi\over {12}}. \end{eqnarray*}$$

17.53   Exercise. A Find the following integrals:

a)

$${\int_0^1 x^2(x^3+1)^3 dx}$$.

b)

$${\int_0^{3/2} {1\over {\sqrt{9-x^2}}}dx}$$.

c)

$${\int_0^1 x\sqrt{1-x}\;dx}$$.

17.54   Exercise. A Find the area of the shaded region, bounded by the ellipse $${ {{x^2}\over 4}+y^2=1}$$ and the lines $x=\pm 1$.

17.55   Example. In practice I would find many of the antiderivatives and integrals discussed in this chapter by computer. For example, using Maple, I would find

> int(sqrt(a^2+x^2),x);

$${\displaystyle \frac {1}{2}} {x} \sqrt {{a}^{2} + {x}^{2}}... ...ln} \left( \! {x} + \sqrt {{a}^{2} + {x}^{2}} \! \right)$$

> int(sin(sqrt(x)),x=0..Pi^2);

$$2 { \pi}$$

> int(sqrt(4 - x^2),x=-1..1);

$$\sqrt {3} + {\displaystyle \frac {2}{3}} { \pi}$$

> int( (sec(x))^3,x);

$${\displaystyle \frac {1}{2}} {\displaystyle \frac {{\rm sin... ...1}{2}} {\rm ln}( {\rm sec}( {x} ) + {\rm tan}( {x} ) )$$

> int(exp(a*x)*cos(b*x),x);

$${\displaystyle \frac {{a} {\rm e}^{( {a} {x} )} {\rm co... ...( {a} {x} )} {\rm sin}( {b} {x} )}{{a}^{2} + {b} ^{2}}}$$